A person finds that the object’s image placed at his normal
near point converges at a distance of 0.5cm behind his retina.Prescribe him a corrective lens of suitable power
The trouble I am having here is that I don't know how far away the corrective lens is from the lens in the eye. I suppose I could simply guess that it is maybe 1 cm? (I'm terrible at estimating distances, so beware!) And even the we need some more information, but I'll give you what I have.
Call the focal length of the lens in the eye f and the focal length of the corrective lens F. I am assuming the object is an infinite distance away from the two lenses. (Or at least the object distance is much larger than either of the focal lengths.)
Without the corrective lens the image forms at a position d, which is 0.5 cm too long. We want it to form at D = d - 0.5 cm. So without the corrective lens we have
$\displaystyle \frac{1}{f} = \frac{1}{d} \implies d = f$
Let's calculate where the image is using both lenses.
We start with the corrective lens:
$\displaystyle \frac{1}{F} = \frac{1}{D'}$
so the image forms at D' = F.
This real image from the corrective lens will be a virtual object for the eye's lens. Thus
$\displaystyle \frac{1}{f} = -\frac{1}{D' - 1} + \frac{1}{D}$
(D being the required image distance, and note that D must be positive for this to be a real image.) We are using D' - 1 for the object because the eye's lens is 1 cm away from the corrective lens.
So
$\displaystyle \frac{1}{f} = -\frac{1}{F - 1} + \frac{1}{d - 0.5}$
or
$\displaystyle \frac{1}{f} = -\frac{1}{F - 1} + \frac{1}{f - 0.5}$
Thus
$\displaystyle F = 2f^2 -f + 1$
Obviously we need to know what f is and I can't even give you an estimate on that.
-Dan